Question

The Fig. shows the position-time graph of a particle of mass 4 kg. What is the:

(a) force on the particle for

t < 0, t > 4s,  0 < t < 4s?

(b) impulse at t = 0 and t = 4s. (consider one dimensional motion only).

Answer 1

(a) For t < 0 and t > 4s, the particle is at rest as the position does not change w.r.t. time. Clearly no force acts on the particle during these intervals.

Further, for D < 1 < 4s, the position of the particle continuously changes with respect to time. As the position-time graph is a straight line, it represents uniform motion and there is no acceleration. Hence, it is also clear that no force acts on the particle during this interval.

(b) Because the velocity is uniform from O to A, hence velocity at O = velocity at A = slope of the graph OA = (3/4) ms^-1

Impulse (at t = 0) = change in momentum

= Final momentum - initial momentum

= 0 - mv = -4(3/4) = -3 kg ms^-1